Ice roof physics question
I have been having some issues with this question. https://isaacphysics.org/questions/ice_roof?board=2ab05fcf-54b6-4bca-aae8-530ac5dd4b10&stage=all
I have calculated the com of the ice bit to be -r*cos... but not sure if it is correct. Any help would be greatly appreciated.
I have calculated the com of the ice bit to be -r*cos... but not sure if it is correct. Any help would be greatly appreciated.
Reply 1
If anybody sees this and is willing to aid, any help would be appreciated.
Reply 2
Original post
by Javier GarcíaIf anybody sees this and is willing to aid, any help would be appreciated.
If anybody is willing to help, it would be greatly appreciated
Reply 3
I know this was a while ago but im struggling on this question as well, do you mind if you could take a look at my working? I have no clue how to get an expression for cos(alpha)
Reply 4
21
Original post
by mosaurlodonI know this was a while ago but im struggling on this question as well, do you mind if you could take a look at my working? I have no clue how to get an expression for cos(alpha)
You need to be a bit more careful calculating the com of the arc. Not sure where your half angles (theta/2) come from and as the question says, the x coordinate of the arc will be
r*sin(theta)
which is the main thing you integrate over from 0 to alpha when you equate the arc's moment to the rod's moment.
Edit - After sleeping on it Id guess your half angles come from the usual expression for the com of an arc? If so, it represents the length and youd have to project it onto the x-axis with another bit of trig and the half angles would combine to give alpha. While this would work if done properly, Id strongly suspect they want you to do it via integration and just concentrate on the x-coordinate above.
Reply 5
Oh I see what you mean - thank you very much 
I might try and see if my original method earlier worked or not and see if the answers are equivalent.
The angles I derived come from the usual expression for the com of an arc but tbh im pretty sure the "half angle" thing I ended up with is wrong - its just the logic that I thought was right.
I might try and see if my original method earlier worked or not and see if the answers are equivalent.
The angles I derived come from the usual expression for the com of an arc but tbh im pretty sure the "half angle" thing I ended up with is wrong - its just the logic that I thought was right.
Reply 6
21
Original post
by mosaurlodonOh I see what you mean - thank you very much
I might try and see if my original method earlier worked or not and see if the answers are equivalent.
The angles I derived come from the usual expression for the com of an arc but tbh im pretty sure the "half angle" thing I ended up with is wrong - its just the logic that I thought was right.
I might try and see if my original method earlier worked or not and see if the answers are equivalent.
The angles I derived come from the usual expression for the com of an arc but tbh im pretty sure the "half angle" thing I ended up with is wrong - its just the logic that I thought was right.
The half angle / usual com of an arc works out, but the total angle is alpha, so the half angles are alpha/2. The com formula gives the length of the "hypotenuse" and you want to use simple trig, so sin(alpha/2) to project it onto the x coordinate. Then just use the double angle formula to get it in terms of cos(alpha).
Reply 7
How would I remove the r*alpha from the top equation?
I substituted my original answer with the equation to check if it was correct and I got x = w/2 which is definitely not right so Ive done something wrong.
I substituted my original answer with the equation to check if it was correct and I got x = w/2 which is definitely not right so Ive done something wrong.
Reply 8
21
Original post
by mosaurlodonHow would I remove the r*alpha from the top equation?
I substituted my original answer with the equation to check if it was correct and I got x = w/2 which is definitely not right so Ive done something wrong.
I substituted my original answer with the equation to check if it was correct and I got x = w/2 which is definitely not right so Ive done something wrong.
Could you write out what youve tried to do? The basic com of an arc will have a sin(alpha/2) and when you project that youll get another multiple which gives a sin^2(alpha/2) which can transformed to cos(alpha) using the usual identity
Reply 9
(ignore post #7)
I've tried to research/watch videos about how to find the basic com of an arc but all I'm finding are the respective coordinates, and the only one that has sin(alpha/2) is y_cm - I thought about using pythagoras or smth but that didnt get me anywhere so I don't actually know the formula for the com of an arc.
Do you derive the com of an arc using x,y coords or is there another way?
I've tried to research/watch videos about how to find the basic com of an arc but all I'm finding are the respective coordinates, and the only one that has sin(alpha/2) is y_cm - I thought about using pythagoras or smth but that didnt get me anywhere so I don't actually know the formula for the com of an arc.
Do you derive the com of an arc using x,y coords or is there another way?
Reply 10
21
Original post
by mosaurlodon(ignore post #7)
I've tried to research/watch videos about how to find the basic com of an arc but all I'm finding are the respective coordinates, and the only one that has sin(alpha/2) is y_cm - I thought about using pythagoras or smth but that didnt get me anywhere so I don't actually know the formula for the com of an arc.Do you derive the com of an arc using x,y coords or is there another way?
I've tried to research/watch videos about how to find the basic com of an arc but all I'm finding are the respective coordinates, and the only one that has sin(alpha/2) is y_cm - I thought about using pythagoras or smth but that didnt get me anywhere so I don't actually know the formula for the com of an arc.
I take it youve done question using the usual approach, so integrate ...
The cog of an arc is
https://www.youtube.com/watch?v=meYuJV1oQRs&ab_channel=MichelvanBiezen
so divide the arc into 2 equal parts and the cog must lie along that line (alpha/2 in this case) and the length is
r sin(alpha/2) / (alpha/2)
in this case. Then project onto the horizontal / x direction and as per the previous post ...
Reply 11
oh yes I see what you mean now - thank you very much.
thats pretty cool you can actually derive that from using coords and doing the way in the video you sent.
you just have to realise sqrt(2-2cosx) = 2sin(x/2) which I did not know.
so it becomes r/alpha * sqrt( sin^2(alpha) + cos^2(alpha) + 1 -2cos(alpha))
= r/alpha * sqrt( 2 - 2cos(alpha))
= r*sin(alpha/2)/alpha/2
thats pretty cool you can actually derive that from using coords and doing the way in the video you sent.
you just have to realise sqrt(2-2cosx) = 2sin(x/2) which I did not know.
so it becomes r/alpha * sqrt( sin^2(alpha) + cos^2(alpha) + 1 -2cos(alpha))
= r/alpha * sqrt( 2 - 2cos(alpha))
= r*sin(alpha/2)/alpha/2
Reply 12
21
Original post
by mosaurlodonoh yes I see what you mean now - thank you very much.
thats pretty cool you can actually derive that from using coords and doing the way in the video you sent.
you just have to realise sqrt(2-2cosx) = 2sin(x/2) which I did not know.
so it becomes r/alpha * sqrt( sin^2(alpha) + cos^2(alpha) + 1 -2cos(alpha))
= r/alpha * sqrt( 2 - 2cos(alpha))
= r*sin(alpha/2)/alpha/2
thats pretty cool you can actually derive that from using coords and doing the way in the video you sent.
you just have to realise sqrt(2-2cosx) = 2sin(x/2) which I did not know.
so it becomes r/alpha * sqrt( sin^2(alpha) + cos^2(alpha) + 1 -2cos(alpha))
= r/alpha * sqrt( 2 - 2cos(alpha))
= r*sin(alpha/2)/alpha/2
WHen you said you didnt know
sqrt(2-2cosx) = 2sin(x/2)
I bet you knew the standard double angle formula
1-2sin^2(x) = cos(2x)
and ...
double and half angle formulae are trivial rearrangements, as are a lot of the trig identities if you try.
(edited 1 year ago)
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